Is there an "easy" way to see that if $\{\phi_n\}_{n=1}^\infty$ is a set of orthonormal functions in a Hilbert space then showing $\langle f, \phi_n \rangle = 0$ for all $n$ implies $f = 0$ is sufficient to show that the orthonormal set is dense?
I know this statement is true because I found it as a theorem in Stein & Sharkarchi, but their proof is fairly involved.
This is more a result about orthogonal complements than orthonormal systems. There are two basic results used here, which are moderately intuitive: (i) $(A^\bot)^\bot = \operatorname{\overline{sp}} A$. (ii) If $D$ is a set whose linear span is dense in $A$, then $x \in A^\bot$ iff $x \in D^\bot$.
Let $S = \operatorname{\overline{sp}} \{ \phi_k \}$. Then $\langle f, \phi_k \rangle = 0$ for all $k$ iff $f \in S^\bot$.
If $\operatorname{{sp}} \{ \phi_k \}$ is dense, then $S = \mathbb{H}$ and so $S^\bot = \{0\}$, hence if $\langle f, \phi_k \rangle = 0$ for all $k$, we have $f \in S^\bot$ and so $f = 0$.
Now suppose that if $\langle f, \phi_k \rangle = 0$ for all $k$, then $f=0$. Then this implies that $S^\bot = \{0\}$. Since $(A^\bot)^\bot = \operatorname{\overline{sp}} A$, we see that $\{0\}^\bot = \mathbb{H} = S$ ($S$ is a closed subspace).